U = 1 b. We fix the identification G a (F ) U sending b to ( 1 b

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1 LECTURE 11: ADMISSIBLE REPRESENTATIONS AND SUPERCUSPIDALS I LECTURE BY CHENG-CHIANG TSAI STANFORD NUMBER THEORY LEARNING SEMINAR JANUARY 10, 2017 NOTES BY DAN DORE AND CHENG-CHIANG TSAI Let L is a global field, G a reductive group over L, and π an irreducible cuspidal automorphic representation of GA L. Recall that this implies that π is an admissible representation and that Flath s theorem says we may decompose π as π = π v with π v an irreducible admissible representation of GL v at least for non-archimedean v, and v running through the places of L. Some properties of π may be read off from the corresponding properties of π v : for example, the levels correspond, and the Hecke eigenvalue a p of π corresponds to the Hecke eigenvalue of π p e.g. when L = Q and G = GL 2. Today, we will restrict our attention to the case of a non-archimedean local field F, and the group G = GL 2 F. Recall that an admissible representation π of G is a complex vector space π with a G-action such that: π = K G K open compact where π K = {v π πk v = v k K}. A representation satisfying this condition is called smooth. π K dim C π K < for any open compact subgroup K G. Inside the group G = GL 2 F, recall that there is a particular subgroup, called the mirabolic subgroup: { } M = a b a F, b F This has a unipotent subgroup { } U = 1 b b F We fix the identification G a F U sending b to 1 b, and think of U as Ga F. Fix a non-trivial additive character ψ : F C. Via the above identification, we think of this as a character on U. If π is an irreducible admissible representation of G, we define: Definition 1. A Whittaker model for π is a G-embedding ι: π Ind G Uψ. Recall that CG is the complex vector space of C-valued functions on G that are invariant by right translation by some open compact subgroup K G. G acts on this space by right translation. We may think of Ind G Uψ as: Likewise, we have: We define: Ind G Uψ = {f CG fug = ψufg u U, g G} Ind M U ψ = {f CM fum = ψufm u U, m M} 1

2 Definition 2. A Kirillov model for π is an M-embedding π Ind M U ψ. Remark 3. Since π might not be irreducible as an M-module, it is no longer obvious that a non-zero M-homomorphism from π to Ind M U ψ is an embedding. Nevertheless, this is true, as we will show later. For f Ind M U ψ we may identify f as a function on F via: fx := f x 0 We may compute the M-action on such f by: a b f x = f x 0 a b = f ax bx = f 1 bx ax 0 = ψbxfax Let KF be the space of functions on F that are left translation-invariant by some 1 + p n F and are supported on a compact subset of F. Here, p is the maximal ideal of O F. Now, we have: Lemma 4. The map given by restricting f Ind M U CM to F 0 defines an isomorphism Ind M U ψ KF. Proof. First, we see that this map lands inside KF : i We have a 0 f x = fax. Since f is right translation-invariant by an open compact subgroup of M, if we take a sufficiently close to 1, this shows fx = fax. i Similarly, fx = 1 b f x = ψbxfx = fx for b p n for some large enough n, i.e. ψbx = 1 for all x suppf, b p n. Suppose ψ is non-trivial on some p m. Then we have suppf p m n+1 is compact. Next, if f KF, we may extend f to an element of Ind M U ψ by defining f a b = ψbfa. The arguments above run in reverse show that f CM because any element of M can be written as a product 1 b a 0 is stable by right translation by an open compact subgroup of M. The same calculations let us easily check that these operations are inverse to each other. Now, Frobenius reciprocity gives: Hom G π, Ind G U ψ = Hom G π, Ind G M Ind M U ψ = Hom M Res G Mπ, Ind M U ψ Thus, if π has a Kirillov model, then π has a Whittaker model since maps from the irreducible G-representation π to Ind G U ψ must be embeddings. We have: Theorem 5. Let π be an irreducible admissible representation of GL 2 that is not one-dimensional. Then π has a Kirrilov model for any fixed choice of ψ. Proof. See [3, ], [1, 4.4], or [2, 3.6]. In addition, the Kirrilov model for π is essentially unique: 2

3 Lemma 6. dim Hom M Res G Mπ, Ind M U ψ = 1 Proof. Suppose we have a Kirrilov model j : π Ind M U ψ KF. Consider the linear functional on π given by L j : f jf1. Since for a F we have jfa = a 0 jf 1 = j a 0 f 1, the functional L j : π C determines j : π KF. Now, note that if f ker L j, then jf1 = 0, so jfa = 0 for all a F sufficiently close to 1, i.e. when va 1 n 0 where v is the valuation on F for some n 0 sufficiently large. Suppose ψ is non-trivial on p m for some m, so that p ψbdb = 0. Now, we have, for all m x F and for all n m + n 0 ψb 1 j π 1 b f x db = ψb 1 ψbxjfx db p n p n = ψbx 1jfx db = 0 p n Now, since j is an embedding of π into a space of functions on F, this says that: I n f := ψb 1 π 1 b f db = 0 p n whenever n n 0. Thus, the kernel of L j is contained in the space of f such that I n f = 0 for all sufficiently large n. We ll see that this is an equality: if jf1 0, then we have: π 1 b f ji n f1 = ψb 1 j p n = ψ0jf1 db p n = jf1 volp n 0 1 db Thus, the kernel of I n is contained in the kernel of L j, so they are equal. This shows that the kernel of L j does not depend on j, so different choices of j can only change L j by a constant multiple. This shows an embedding j : Res G Mπ Ind M U ψ is unique up to a constant. But it is not a priori clear that a non-trivial homomorphism j from Res G Mπ to Ind M U ψ is injective. Choose some such j, and let j be an embedding of Res G Mπ into Ind M U ψ. Since j is non-trivial, we see that L j 0. Now, if f ker L j = {f I n f = 0 n 0}, the argument above shows that L j f = 0, so ker L j ker L j. Since these are both codimension-one subspaces of π, we must have ker L j = ker L j and thus L j = cl j for some c 0. As before, this implies that j = cj for some c 0 and we see that j is an embedding after all. Let us now fix an irreducible admissible representation π of dimension greater than 1, and fix a Kirillov model j : π Ind M U ψ. We will drop j from the notation and regard π as a subspace of Ind M U ψ. Define SF KF to be the subspace of functions that are compactly supported in F, i.e. the elements of KF that vanish in a neighborhood of 0. We call these Schwarz functions on F. 3

4 Lemma 7. SF π. Proof. sketch First, we verify that SF is an irreducible representation of M. This can be done as follows: Suppose ψ is trivial on p m+1 but not on p m. Let f SF be non-trivial, and suppf a + p n for some a F. Then we have 1 ψba 1 ψbxfxdb = p m n p m n { fx if x a + p n+1 0 else In other words, f p m n ψba 1 1 b fdb restricts f to a + p n+1 suppf. By using this, one can construct from any non-zero function in SF any characteristic on F and thus any function in SF. Then, we have π SF 0, because π 1 b f f SF for any b F. We claim that we may choose b, f such that this is non-zero. This is because: π 1 b f f x = ψbx 1 fx Since ψ is non-trivial, this is not always equal to 0. Since SF is irreducible, this means that SF π. This allows us to state the next lemma: Lemma 8. Let w = 1 0. Then we have: π = SF + πwsf Proof. Since π is an irreducible G-representation, π is spanned by πg SF for g G. Now, we may use the Bruhat decomposition to write G = B UwB for B = M Z the upper triangular Borel subgroup of G Z is the center of G = GL 2 F, i.e. the scalar matrices. Thus, π is generated by SF and πuwsf for u U. But we have: πuwf πwf = πu 1πwf SF as we may see by writing u = 1 b and using the explicit description of the action of M on KF. We will need the following general definition: Definition 9. Let π be a smooth representation of G. The contragredient representation ˇπ is the space of smooth functionals on π, i.e. we have: ˇπ = {l: π C compact open K such that lπkf = lf f π} In other words, ˇπ is the subspace of smooth vectors in the linear dual of π. If π is admissible, then for all K, π = π K π K, where π K is the kernel of e K = 1 µk 1 K HG, K, since e K acts on π as projection onto π K. This implies that the linear dual of π splits as π = π K π K, so π K = π K. Since π K = ˇπ K, this shows that ˇπ K is 4

5 finite-dimensional, and thus ˇπ is admissible. Moreover, the natural pairing, : π ˇπ C is non-degenerate, as it restricts to a non-degenerate pairing on π K ˇπ K. Additionally, this shows that the natural map π ˇπ is an isomorphism, since both sides are the sum of K-fixed subspaces, yet the map restricts to an isomorphism π K = ˇπ K for each K. Let us also remark that if π, π are admissible representations, and there exists a non-degenerate pairing, : π π C such that πgf, π gf = f, f for all f π, f π, g G, then the natural map π ˇπ is an isomorphism. Non-degeneracy implies that the map is injective, and we can check surjectivity by passing to π K, ˇπ K, and comparing dimensions. Now, fix an irreducible admissible representation π and a Kirillov model π KF. Recall that π 0 c 0 c acts by a constant by Schur s lemma. Thus, we may write: We have: π c 0 0 c = w π c id π Theorem 10. The contragredient representation ˇπ of π can be realized in the same underlying abstract vector space as π with the action ˇπg = w π det g 1 πg. Proof. See [3, 1.6]. Then, a Kirillov model ǰ for ˇπ is given by ǰfx = w π x 1 jfx. This gives us an embedding j ǰ : π ˇπ KF KF. This lets us define a pairing on π ˇπ by: f, ˇf := f 1 x ˇf x d x + F f 2 xˇπ ˇf x d x F Here, we choose f 1, f 2 SF such that f = f 1 + πwf 2. We may define a pairing, for f, f π: f, f = f 1 xw π x 1 f x d x + F f 2 xw π x 1 πwf x d x F Thus, the essential content of Theorem 10 says that this pairing is non-degenerate, well-defined i.e. it does not depend on the choice of representation of f in terms of f 1, f 2, and that: Now, we may define: πgf, πgf = w π det gf, f Definition 11. An irreducible admissible representation π of G is called supercuspidal if for all f π, ˇf ˇπ, the matrix coefficient g πgf, f is a compactly supported function on G mod the center Z of G. Note that this definition makes sense for any reductive group G, since the definition of ˇπ of the pairing between π and ˇπ are both perfectly general. We have the following theorem: Theorem 12. π is supercuspidal iff π = SF in the Kirillov model. 5

6 Remark 13. This theorem does not yet show that any supercuspidal representations exist! Note that in general SF is only a M-subrepresentation of π. Proof. First, suppose that π = SF. Note that by the previous discussion, this implies that ˇπ = SF as well, since they have the same underlying space and essentially the same Kirillov model up to a central character, which does not affect the condition of being compactly supported away from 0. Then for all f π, ˇf ˇπ, we have: π a 0 f, ˇf = F fax ˇf x d x Since f, ˇf SF, we have ˇf x = 0 for x outside a compact subset of F. Thus since inversion and multiplication are continuous on F, Supp ˇf 1 Suppf is a compact subset of F and the integrand vanishes for a outside this set. Recall that we have the Cartan decomposition: G = a F K a 0 K Z for K = GL 2 O F. Fix f π, ˇf ˇπ, and let K 1 K be such that K 1 fixes both f and ˇf. Since [K : K 1 ] <, the K-orbit of f is finite and similarly for ˇf. Let f 1,..., f s, ˇf 1,..., ˇf r be the elements of these orbits. We see that there is a compact subset Ω F such that π a 0 f i, ˇf j = 0 for all i, j and any a F Ω. Then, for any g, we may write: πgf, ˇf = πk 1 a 0 k 2 zf, ˇf = w π z π a 0 πk 2 f, πk1 1 ˇf = w π z π a 0 f i, ˇf j Thus, this is 0 unless g K Ω 0 K Z, and this set is compact mod Z. Therefore, π is supercuspidal. Conversely, suppose that π is supercuspidal. Take any ˇf ˇπ. For any f SF π and a F we again have: π a 0 f, ˇf = fax ˇf xd x F and we know that as a function of a, this is compactly supported in F. As ˇf is smooth, as a function on F it is locally constant under some 1 + p n. By taking f = 1 1+p n, the above compact support property implies that ˇf SF. Thus, ˇπ = SF and therefore we know that π = SF by the relationship between the Kirillov models of π and ˇπ. 6

7 References [1] D. Bump, Automorphic forms and representations, Cambridge Univ. Press, Cambridge, [2] Colin J. Bushnell and Guy Henniart, The local Langlands conjecture for GL2, Springer-Verlag, Berlin, [3] R. Godement, Notes on Jacquet Langlands Theory, IAS lecture notes, Princeton, Online at stanford.edu/ conrad/conversesem/refs/godement-ias.pdf. 7